Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
int n;
int tx[11111];
int x[11111];
int y[11111];
int ty[11111];
int base[33];
bool u[11111];
bool uy[11111];
bool check(vector<pair<int, int> > v) {
  for (int i = (0); i < (v.size()); i++) {
    tx[v[i].first] ^= tx[v[i].second];
  }
  for (int i = (0); i < (n); i++)
    if (tx[i] != ty[i]) {
      puts("FUCK");
      while (1)
        ;
      return false;
    }
  return true;
}
int main() {
  cin >> n;
  for (int i = (0); i < (n); i++) scanf("%d", x + i);
  for (int i = (0); i < (n); i++) scanf("%d", y + i);
  memset(base, -1, sizeof base);
  memset(u, false, sizeof u);
  memset(uy, false, sizeof uy);
  for (int i = (0); i < (n); i++) tx[i] = x[i];
  for (int i = (0); i < (n); i++) ty[i] = y[i];
  vector<pair<int, int> > v;
  for (int bit = 0; bit < 30; bit++) {
    for (int i = (0); i < (n); i++) {
      if ((x[i] & (1 << bit)) && !u[i]) {
        u[i] = true;
        base[bit] = i;
        for (int j = (0); j < (n); j++)
          if (j != i)
            if (x[j] & (1 << bit)) {
              x[j] ^= x[i];
              v.push_back(pair<int, int>(j, i));
            }
        break;
      }
    }
  }
  bool good = true;
  for (int i = (0); i < (n); i++)
    if (!u[i]) {
      while (x[i])
        ;
      for (int j = (0); j < (30); j++)
        if (base[j] != -1) {
          if (y[i] & (1 << j)) {
            x[i] ^= x[base[j]];
            v.push_back(pair<int, int>(i, base[j]));
          }
        }
      if (x[i] != y[i]) {
        good = false;
        break;
      }
    }
  if (good) {
    vector<int> pos;
    for (int i = (0); i < (n); i++)
      if (u[i]) pos.push_back(i);
    vector<pair<int, int> > inv;
    for (int bit = 0; bit < 30; bit++) {
      for (int i = (0); i < (pos.size()); i++) {
        if ((y[pos[i]] & (1 << bit)) && !uy[pos[i]]) {
          uy[pos[i]] = true;
          for (int j = (0); j < (pos.size()); j++)
            if (j != i)
              if (y[pos[j]] & (1 << bit)) {
                y[pos[j]] ^= y[pos[i]];
                inv.push_back(pair<int, int>(pos[j], pos[i]));
              }
          break;
        }
      }
    }
    for (int i = (0); i < (pos.size()); i++) {
      if (y[pos[i]] == 0) continue;
      bool f = 0;
      for (int bit = (0); bit < (30); bit++)
        if (base[bit] != -1 && (y[pos[i]] & (1 << bit))) {
          f = 1;
          if (base[bit] != pos[i]) {
            for (int bit2 = (0); bit2 < (30); bit2++)
              if (base[bit2] == pos[i]) {
                v.push_back(pair<int, int>(pos[i], base[bit]));
                v.push_back(pair<int, int>(base[bit], pos[i]));
                v.push_back(pair<int, int>(pos[i], base[bit]));
                swap(x[pos[i]], x[base[bit]]);
                swap(base[bit], base[bit2]);
                break;
              }
          }
          for (int bit2 = (0); bit2 < (30); bit2++)
            if (base[bit2] != -1 && ((y[pos[i]] ^ x[pos[i]]) & (1 << bit2))) {
              x[pos[i]] ^= x[base[bit2]];
              v.push_back(pair<int, int>(pos[i], base[bit2]));
            }
          break;
        }
      if (!f) {
        good = 0;
        break;
      }
      if (x[pos[i]] != y[pos[i]]) {
        good = 0;
        break;
      }
    }
    for (int i = (0); i < (pos.size()); i++) {
      if (y[pos[i]]) continue;
      inv.push_back(pair<int, int>(pos[i], pos[i]));
    }
    reverse(inv.begin(), inv.end());
    for (int i = (0); i < (inv.size()); i++) {
      v.push_back(inv[i]);
    }
  }
  if (!good) {
    puts("-1");
  } else {
    cout << v.size() << endl;
    for (int i = (0); i < (v.size()); i++)
      printf("%d %d\n", v[i].first + 1, v[i].second + 1);
  }
  return 0;
}